I just need help with d e and f plz d) Evaluate the limit lim msec(x).x5

Answered: d) Evaluate the limit lim msec(x). x5

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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I just need help with d e and f plz

Expert Solution

Step 1

In the given question m_secdenotes the slope of the secant line joining the points (5, f(5)) and (x, f(x)) for both of the pieces of the function f(x)

Step 2

$f (x) = \{\begin{cases} \frac{x^{2}}{4} - x + 1 0 \leq x \leq 5 \\ 3 \sqrt{5} \sqrt{x} - \frac{51}{4} 5 < x \leq 10 \end{cases}$

so $m_{s e c} (x) = \frac{f (x) - f (5)}{x - 5}$

For, $0 \leq x \leq 5$ , the limit of the secant $m_s e c (x)$ is

$l i m_{x \to 5^{-}} m_{s e x} (x) = l i m_{x \to 5^{-}} \frac{f (x) - f (5)}{x - 5}$

Step 3

$l i m_{x \to 5^{-}} m_{s e x} (x) = l i m_{x \to 5^{-}} \frac{(\frac{x^{2}}{4} - x + 1) - (\frac{5^{2}}{4} - 5 + 1))}{x - 5}$

$l i m_{x \to 5^{-}} m_{s e x} (x) = l i m_{x \to 5^{-}} \frac{\frac{x^{2} - 5^{2}}{4} - (x - 5)}{x - 5}$

$l i m_{x \to 5^{-}} m_{s e x} (x) = l i m_{x \to 5^{-}} \frac{(x - 5) (x + 1)}{4 (x - 5)}$

$l i m_{x \to 5^{-}} m_{s e x} (x) = l i m_{x \to 5^{-}} \frac{(x + 1)}{4}$

$l i m_{x \to 5^{-}} m_{s e x} (x) = \frac{6}{4}$

$l i m_{x \to 5^{-}} m_{s e x} (x) = \frac{3}{2}$

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